Solve inequality involving modulus

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    Inequality Modulus
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Homework Help Overview

The discussion revolves around solving the inequality involving a modulus expression: \large \frac{|x-1|(x-3)(x-5)^{2010}}{(|x|-3)(|x|+1)} \geq 0. Participants are exploring the implications of critical points and the behavior of the modulus terms within the inequality.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss identifying critical points, noting 3 and 5, while also questioning how to handle the modulus expressions. There are inquiries about the specific values of x that make the modulus terms equal to zero.

Discussion Status

The discussion is active with participants sharing their thoughts on critical points and the effects of modulus. Some suggest breaking the problem into ranges defined by critical points, while others highlight the need for simplification before proceeding.

Contextual Notes

Participants are required to use the Wavy-Curve method for solving the inequality. There is a focus on understanding the behavior of the modulus terms, particularly in relation to the critical points identified.

utkarshakash
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Homework Statement


Find the solution set of [itex]\large \frac{|x-1|(x-3)(x-5)^{2010}}{(|x|-3)(|x|+1)} \geq 0[/itex]

Homework Equations


I am required to solve this using Wavy-Curve method

The Attempt at a Solution


The critical points are 3 and 5. But I don't know what to do with expressions involving modulus signs.
 
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utkarshakash said:

Homework Statement


Find the solution set of [itex]\large \frac{|x-1|(x-3)(x-5)^{2010}}{(|x|-3)(|x|+1)} \geq 0[/itex]

Homework Equations


I am required to solve this using Wavy-Curve method

The Attempt at a Solution


The critical points are 3 and 5. But I don't know what to do with expressions involving modulus signs.
For what values of x is |x|-3 = 0 ?

For what values of x is |x|+1 = 0 ?
 
utkarshakash said:
The critical points are 3 and 5. But I don't know what to do with expressions involving modulus signs.
Not sure how you're defining critical points, but interesting things will happen at -3, 1, 3 and 5. I would break it into the five ranges those points generate and consider each separately.
But first, there's a very easy simplification. Think about the (|x|+1) term.
 
SammyS said:
For what values of x is |x|-3 = 0 ?

For what values of x is |x|+1 = 0 ?

Okay I got the answer.
 

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